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Given 2.25 moles of Al, and 3.25 moles of Cl2 in a synthesis reaction, answer the following questions.?
Determine the percent yield if 2.00 moles of AlCl3 was collected during the experiment.
How much excess reactant was left over?
1 Answer
- Roger the MoleLv 72 months ago
2 Al + 3 Cl2 → 2 AlCl3
3.25 moles of Cl2 would react completely with 3.25 x (2/3) = 2.17 moles of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.
(3.25 mol Cl2) x (2 mol AlCl3 / 3 mol Cl2) = 2.1667 mol AlCl3 in theory
(2.00 mol) / (2.1667 mol) = 0.923 = 92.3% yield AlCl3
((2.25 mol Al initially) - (2.17 mol Al reacted)) x (133.3405 g AlCl3/mol) =
10.7 g Al left over
